Averages, Mixtures & Alligations: Aptitude Questions with Answers
Review Averages, Mixture and Alligations concepts
Q. 1 In a lively classroom of five students, the average age is 20 years. One day, a student who is 24 years old leaves the group and is replaced by a new student. Curiously, this change brings down the average age of the group to 19 years. Can you determine the age of the newcomer?
A) 16 years
B) 15 years
C) 14 years
D) 18 years
Check Solution
Ans: D) 18 years
Total age of 5 students initially = $5 \times 20 = 100$
After replacing the student aged 24, total age of 5 students = $5 \times 19 = 95$
Age of the new student = 95−(100−24)=95−76=19
Q. 2 The average of 20 numbers is zero. Out of these, at most, how many numbers can be greater than zero?
A) 19
B) 20
C) 10
D) 1
Check Solution
Ans: A) 19
For the average to be zero, the sum of positive numbers must equal the sum of negative numbers.
So, at most, 19 numbers can be positive if only 1 number is negative to balance their sum.
Q. 3 A farmer has a large pot holding 60 liters of a milk-and-water mixture. Initially, the ratio of milk to water is 2:1. To adjust the composition for his recipe, he adds some water, changing the ratio to 1:2. How much water does the farmer add?
A) 40 l
B) 30 l
C) 20 l
D) 60 l
Check Solution
Ans: A) 40 l
Amount of milk = $\frac{2}{3} \times 60 = 40$
Amount of water = $\frac{1}{3} \times 60 = 20$
Let x liters of water be added to make the ratio 1:2. Then,
$\frac{40}{20 + x} = \frac{1}{2}$
Solving, x=60−20=40
Q. 4 Imagine a cyclist embarking on an adventure! She covers the first half of a certain distance at a steady speed of 30 km/hr and picks up pace in the second half, traveling at 50 km/hr. What would her average speed be for the entire journey?
A) 37.5 km/hr
B) 40 km/hr
C) 45 km/hr
D) 42 km/hr
Check Solution
Ans: A) 37.5 km/hr
Let the total distance be 2D.
Time taken for first half = $\frac{D}{30}$
Time taken for second half = $\frac{D}{50}$
Total time = $150\frac{D}{30} + \frac{D}{50} = \frac{5D + 3D}{150} = \frac{8D}{150}$
Average speed = $\frac{2D}{8D/150} = \frac{2 \times 150}{8} = 37.58$ km/hr
Q. 5 Eight friends decide to track their weight as part of a health challenge. Their average weight increases by 2.5 kg when one of them, who weighs 70 kg, is replaced by a new person. What is the weight of the new person?
A) 90 kg
B) 88 kg
C) 85 kg
D) 80 kg
Check Solution
Ans: A) 90 kg
Increase in total weight = $2.5 \times 8 = 20$ kg
Weight of the new person = 70 + 20 = 90 kg
Q. 6 A mysterious jar holds a mix of two liquids, A and B, in the ratio 4:3. A mischievous scientist removes 10 liters of the mixture and replaces it with pure liquid B. The new ratio becomes 2:3. What is the original volume of the mixture?
A) 35 liters
B) 40 liters
C) 42 liters
D) 45 liters
Check Solution
Ans: A) 35 liters
Let the original quantity of mixture be x liters.
Amount of A = $\frac{4}{7}x$ , Amount of B = $\frac{3}{7}x$
After replacing 10 liters,
New amount of A = $\frac{4}{7}x – \frac{4}{7} \times 10 = \frac{4x – 40}{7}$
New amount of B = $\frac{3}{7}x – \frac{3}{7} \times 10 + 10 = \frac{3x + 40}{7}$
Equating the ratio,
$\frac{\frac{4x – 40}{7}}{\frac{3x + 40}{7}} = \frac{2}{3}$
Solving for x , we get x=35
Q. 7 The average age of a school group of 15 children is 10 years. Soon, they are joined by five younger kids whose average age is 8 years. What becomes the new average age of this group?
A) 9.5 years
B) 9 years
C) 10 years
D) 8.5 years
Check Solution
Ans: A) 9.5 years
Total age of the 15 children = $15 \times 10 = 150$
Total age of the 5 new children = $5 \times 8 = 40$
New total age = 150+40=190
New average age = $9.5\frac{190}{20} = 9.5$ years
Q. 8 Two types of rice are blended to create the perfect mixture for a chef’s special dish. One costs Rs. 60 per kg, and the other costs Rs. 90 per kg. They are mixed in the ratio 2:3. What is the cost per kg of this exquisite mixture?
A) Rs. 75
B) Rs. 80
C) Rs. 78
D) Rs. 85
Check Solution
Ans: A) Rs. 75
Using the alligation rule: Price of mixture=$\frac{2 \times 90 + 3 \times 60}{2 + 3} = \frac{180 + 180}{5} = 75$
Q. 9 A class of 50 students has an average score of 72 in a challenging exam. After removing the scores of the top five performers, the remaining 45 students have an average score of 70. Can you determine the average score of the top five students?
A) 90
B) 82
C) 80
D) 85
Check Solution
Ans: A) 90
Total score of 50 students = $50 \times 72 = 3600$
Total score of remaining 45 students = $45 \times 70 = 3150$
Sum of top 5 scores = 3600 – 3150 = 450
Average score of top 5 students = $\frac{450}{5} = 90$
Q. 10 A barrel contains a flavourful mixture of alcohol and water in the ratio of 3:1. After adding 10 liters of water, the ratio transforms to 2:3. How much alcohol was present in the barrel initially?
A) 6 liters
B) 8 liters
C) 9 liters
D) 10 liters
Check Solution
Ans: C) 9 liters
Let the initial quantity of alcohol be 3x and water x.
After adding 10 liters of water,
New ratio = $\frac{3x}{x + 10} = \frac{2}{3}$
Solving, x=9, so the initial amount of alcohol = $73 \times 9 = 27$
Q. 11 Find the average of following values : 710, 790, 900, 680, 300
A) 674
B) 676
C) 675
D) 677
Check Solution
Ans: B
Find sum of the numbers first. Sum = 2670.
Divide the sum by 5 which is the total count. Average = 2670 / 5 = 676
Q. 12 Find the average of following values : 410, 890, 280, 640, 310
A) 507
B) 504
C) 505
D) 506
Check Solution
Ans: D
Find sum of the numbers first. Sum = 2120.
Divide the sum by 5 which is the total count. Average = 2120 / 5 = 506
Q. 13 Find x, if average of the quantities is 80. Following are the quantities : 94, 89, 36, 82, x
A) 98
B) 99
C) 97
D) 100
Check Solution
Ans: B
Sum of all the quantities is equal to average multiplied by number of quantities. Sum = 80 * 5 = 400. Now this sum = 94 + 89 + 36 + 82 + x . Solving it, we will get value of unkown quantity = 99
Q. 14 A mechanic repaired five cars, and the repair times were 24, 88, 49, 88, and 33 hours. What was the average repair time?
A) 55.4
B) 54.4
C) 56.4
D) 57.4
Check Solution
Ans: C
Find sum of the values first. Sum = 258.
Divide the sum by 5 which is the total count. Average = 258 / 5 = 56.4
Q. 15 A tank is filled with a solution totaling 10,000 gallons, where 5% of it is sodium chloride. If 2,500 gallons of the solution evaporate, what will be the new concentration percentage of sodium chloride in the remaining solution?
A) $ 1.25\% $
B) $ 3.75\% $
C) $ 6.25\% $
D) $ 6.67\% $
Check Solution
Ans: D
The tank contains 10,000 gallons of solution, of which 5% is sodium chloride, so there are 500 gallons of sodium chloride in the solution. If 2,500 gallons of water evaporate, the remaining solution will be 7,500 gallons, and the concentration of sodium chloride will be 500 / 7,500 = 6.67%.
Q. 16 There are eight liters of a solution that contains 45% salt. After some of the water in this solution evaporates while being exposed to sunlight, the concentration of salt increases to 60%. What is the approximate volume of water that has evaporated?
A) 1.2 liters
B) 2.0 liters
C) 2.1 liters
D) 6.0 liters
Check Solution
Ans: B
Let x be the amount of water evaporated from the solution.
Initially, there are 8 liters of a 45% saline solution, which means there are 0.45*8 = 3.6 liters of saline and 4.4 liters of water.
After the water evaporates, there are 8-x liters of solution left. Since the solution is now 60% saline, there are 0.6*(8-x) = 4.8-0.6x liters of saline.
We know that the amount of saline remains constant, so we can set up the following equation:
3.6 = 4.8-0.6x
Solving for x, we get:
x = 2 liters
Therefore, approximately 2 liters of water evaporated from the solution.
Q. 17 How much water must be added to 10 gallons of $ 10\% $ brine solution to decrease the concentration to 7 % ?
A) 0-1.5 gal
B) 1.5-3 gal
C) 3-4.5 gal
D) 4.5-6 gal
Check Solution
Ans: C
Let amount of water that must be added in gallons to dilute concentration of brine solution = x
Amount of brine already present = 10/100 * 10 = 1 gallon
1/(10+x) *100 = 7
=>100 = 70 + 7x
=> x= 30/7 = 4.2
Q. 18 Two copper alloys have varying copper contents. Alloy A weighs 6 kg, and Alloy B weighs 12 kg. An identical amount of each alloy was removed and exchanged; the removed portion from Alloy A was mixed with Alloy B, and the removed portion from Alloy B was mixed with Alloy A. The mixing resulted in two new alloys, both having the same percentage of copper. How much weight was removed from each alloy?
A) 1 kg
B) 2 kg
C) 3 kg
D) 4 kg
Check Solution
Ans: D
Let $x$ be the weight of the piece that was removed. The first alloy has $6x/6=x$ kilograms of copper, and the second alloy has $12x/12=x$ kilograms of copper. After the pieces were removed, the first alloy had $6-x$ kilograms of copper and the second alloy had $12-x$ kilograms of copper. The new alloys have the same percentage of copper, so the ratio of copper to total weight is the same in both alloys. This means that $\frac{x}{6-x}=\frac{x}{12-x}.$ Solving for $x$, we get $x=4$. Therefore, the weight of the piece that was removed is $\boxed{4}$ kilograms.
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